# Help showing the classical Legendre equation has limit circle boundary points.

I am following this paper by Krall and Zettl. I am trying to use the results of Sturm-Louiville (SL) theory to study eigen functions of the classical Legendre equation:

$$\tag1 \frac{d}{dx}\left((1-x^2)\frac{du}{dx}\right)=\lambda u$$

This SL problem is singular at the points $$x=\pm1$$ as around these points $$(1-x^2)^{-1}$$ is not locally integrable. I am trying to show that the end points are limit-circle. i.e. that for all solutions $$u$$, of $$(1)$$ that

$$\tag 2u\in L^2(-1,\beta) \quad \forall \beta \in (-1,1) \quad\text{ ( so x=-1 is limit circle)}$$

$$\tag 3u\in L^2(\alpha,1) \quad \forall \alpha \in (-1,1) \quad\text{ ( so x=1 is limit circle)}$$

I know the solutions of $$(1)$$ are given by the Legendre functions $$P_\lambda(x),Q_\lambda(x)$$ and so I think showing $$(2)$$ and $$(3)$$ is equivalent to showing that the $$L^2$$ norms for both (??) $$P_\lambda(x)$$ and $$Q_\lambda(x)$$ are finite, i.e. showing

$$\int^\beta_{-1} P_\lambda(x)^2 dx, \int^\beta_{-1} Q_\lambda(x)^2 dx \quad\text{and}\quad\int^1_{\alpha} P_\lambda(x)^2 dx,\int^1_{\alpha} Q_\lambda(x)^2 dx$$ are finite.

Here is where I get confused, it is stated in the above linked paper that the SL problem is indeed limit circle at both endpoints.

But,

• If $$\lambda$$ is taken to be an integer then $$P_\lambda(x)$$ is a finite degree polynomial and $$||P_\lambda||<\infty$$ is obvious. However $$Q_\lambda(x)$$ can still be singular in this case... Is it the case that $$Q_\lambda(x)$$ diverges slow enough as $$x \to \pm1$$ to have a finite $$L^2$$ norm?
• In the case $$\lambda$$ is not an integer (this case I am most interested in), the behaviour of $$P_\lambda$$ and $$Q_\lambda$$ only get more singular around $$x=\pm1$$.

In the linked article by Krall et.al. it is mentioned that the classification of boundary points into limit circle / limit point is independent of $$\lambda$$ so I am assuming there must be some argument in the non-integer $$\lambda$$ case which shows the poles of $$P_\lambda$$ and $$Q_\lambda$$ diverge sufficiently slowly to be in $$L^2$$.

My first question is how can I show this? I suspect it might be possible by considering the representations of $$P_\lambda$$ and $$Q_\lambda$$ in terms of the hypergeometric series, however I have little experience with this and am worried about $$_2F_1$$ diverging when $$x=-1$$ since I think this is outside of $$_2F_1$$'s radius of convergence... Maybe there is an asymptotic argument?

My second question concerns the study of the associated Legendre equation, where the solutions are given by the associated Legendre functions $$P^\mu_\lambda(x)$$,$$Q^\mu_\lambda(x)$$.

In this case are the boundary points still limit circle? To show it I think there might be a different weight function $$w(x)$$ but I think the procedure should be more or less then same.

Would the spectrum still be discrete for the non-integer associated Legendre equation? (for example when the boundary conditions do not require bounded solution at the end points but only $$L^2$$)

In order to show that the Legendre equation is in the limit circle case, take a look at the solutions of $$\frac{d}{dx}\left((1-x^2)\frac{df}{dx}\right)=0.$$ This has explicit, classical solutions involving two constants, $$A$$, $$B$$: $$(1-x^2)\frac{df}{dx} = A \\ \frac{df}{dx} = A\frac{1}{1-x^2}=\frac{A}{2}\left[\frac{1}{1-x}+\frac{1}{1+x}\right] \\ f = A\left[-\frac{1}{2}\ln(1-x)+\frac{1}{2}\ln(1+x)\right]+B$$ All solutions are square integrable on $$(-1,1)$$, regardless of the choice of $$A$$ and $$B$$. So the Legendre equation is in the limit-circle case at both $$x=-1$$ and $$x=1$$. That means that the classical Legendre equation requires an endpoint condition at $$x=-1$$ and at $$x=1$$. It turns out that requiring boundedness near $$x=\pm 1$$ gives a Sturm-Liouville problem where the only eigenfunctions are the classical Legendre polynomials.
• Oh yes, since the LC classification is independent of $\lambda$ we should just take it equal to zero to get the much simpler equation in your answer and solve that. Thanks for your answer! Nov 21, 2021 at 9:29
• @valcofadden : Yes, that is correct. And you can find a basis of endpoint functionals that are continuous on the graph of $L$ by using the null space vectors $f=A\left[\cdots\right]+B$ shown above. These endpoint functionals determine all possible self-adjoint versions of the Legendre operator $Lg=-\frac{d}{dx}\left((1-x^2)\frac{dg}{dx}\right)$ in terms of endpoint conditions on $g$ at $\pm 1$. Nov 22, 2021 at 5:29